TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-26

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For every prime power $q$ , the space $P G (2, q)$ is a projective plane of order $q$

Statement

Let $q$ be a prime power, choose a field $F$ with $q$ elements, and consider the incidence structure whose points are the one-dimensional linear subspaces of $F^{3}$ , whose lines are the two-dimensional linear subspaces of $F^{3}$ , and where incidence is inclusion. Then this structure is a finite projective plane of order $q$ . It is denoted $P G (2, q)$ .

Facts & Assumptions

Given: A prime power $q$ .

[L1]

There exists a field $F$ with exactly $q$ elements (For every prime $p$ and $n \geq 1$ , a field with $p^{n}$ elements exists).

[L2]

A one-dimensional $F$ -vector space has $q$ elements, a two-dimensional one has $q^{2}$ elements, and $F^{3}$ has $q^{3}$ elements (A $d$ -dimensional vector space over a field with $q$ elements has exactly $q^{d}$ elements).

Proof

technique · direct

1.1

Choose a field $F$ with $q$ elements by [L1], and let points and lines be the one-dimensional and two-dimensional subspaces of $F^{3}$ .

L1choose

2.1

If $U$ and $V$ are distinct points, choose nonzero vectors $u \in U$ and $v \in V$ . They are linearly independent, so their span is a two-dimensional subspace containing both $U$ and $V$ . Any two-dimensional subspace containing $U$ and $V$ contains $u$ and $v$ , hence contains their span, so this line is unique.

step 1.1algebra

2.2

Let $W_{1}$ and $W_{2}$ be distinct lines. Each has $q^{2}$ elements by [L2]. If $W_{1} \cap W_{2} = {0}$ , then the map $W_{1} \times W_{2} \to F^{3}$ , $(x, y) \mapsto x + y$ , is injective, so $F^{3}$ would have at least $q^{4}$ elements, contradicting [L2]. Thus $W_{1} \cap W_{2}$ contains a nonzero vector and therefore at least one point. If it contained two distinct points, then it would contain the two-dimensional span of those points, forcing $W_{1} = W_{2}$ . So distinct lines meet in exactly one point.

step 1.1L2algebra

2.3

The four one-dimensional subspaces $⟨ e_{1} ⟩$ , $⟨ e_{2} ⟩$ , $⟨ e_{3} ⟩$ , and $⟨ e_{1} + e_{2} + e_{3} ⟩$ have no three on one line: the span of any two coordinate axes is the set of vectors with one coordinate $0$ , which does not contain $e_{1} + e_{2} + e_{3}$ , and the span of $⟨ e_{i} ⟩$ with $⟨ e_{1} + e_{2} + e_{3} ⟩$ does not contain either of the other two coordinate axes.

step 1.1algebra

2.4

Let $W$ be a line. By [L2], $W$ has $q^{2} - 1$ nonzero vectors. Each point on $W$ has $q - 1$ nonzero vectors, and two distinct points meet only in $0$ , so the points on $W$ partition the nonzero vectors of $W$ into pieces of size $q - 1$ . Hence $W$ contains $(q^{2} - 1) / (q - 1) = q + 1$ points, which is at least $3$ because every prime power satisfies $q \geq 2$ .

step 1.1L2algebra

3.1

Steps 2.1, 2.2, 2.3, and 2.4 verify the axioms of A finite projective plane, and step 2.4 identifies the common line size as $q + 1$ . Therefore The order of a finite projective plane gives order $q$ .

step 2.1step 2.2step 2.3step 2.4∎

Depends on

Used by

PG(2,3) has 13 points and 13 lines Example

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17 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Noam D. Elkies, Math 155 notes: Feb. 3 (standard reference, not scraped)
K. Conrad, Finite Fields, Sections 1-2 (standard reference, not scraped)

For every prime power qq, the space PG(2,q)PG(2,q) is a projective plane of order qq